You swing a weight attached to a string in a vertical circle. At the top of the circle the string just barely goes slack for an instant. At this point the centripetal acceleration of the weight is

a) Greater than g.
b) Less than g.
c) Equal to g.
d) Can’t answer without knowing the weight’s speed.

When the string just barely goes slack at the top of the circle, it means that the tension in the string has dropped to zero for an instant, causing the weight to stop moving in a circular path. At this point, the centripetal acceleration of the weight is equal to the acceleration due to gravity (g).

Centripetal acceleration is given by the formula a = v^2 / r, where v is the speed of the object and r is the radius of the circular path. At the top of the circle, the speed of the weight can be calculated using the conservation of energy, where gravitational potential energy becomes kinetic energy. At the top, the weight's velocity is momentarily zero and its kinetic energy becomes zero.

Now, assuming the speed of the weight just before the string goes slack is v, the centripetal acceleration is a = v^2 / r. At that instant, the weight is at rest. This centripetal acceleration is solely due to the force of gravity acting as the centripetal force when the string goes slack. Since no other force is acting on the particle, the centripetal acceleration due to gravity is g.

So, when the string goes slack at the top of the circle, the centripetal acceleration of the weight is when it equal to g.

msm555 msm555 · Answer 2 · 2024-01-06T08:41:05+01:00

Answer:

c) Equal to g.

Explanation:

When the string just barely goes slack at the top of the circle, it means that the tension in the string has become zero. At this instant, the centripetal force acting on the weight is provided solely by gravity.

The centripetal force ([tex]\sf F_c[/tex]) required to keep an object moving in a circular path is given by:

[tex]\sf F_c = \dfrac{m \cdot v^2}{r} [/tex]

where:

[tex]\sf m[/tex] is the mass of the object,
[tex]\sf v[/tex] is the speed of the object,
[tex]\sf r[/tex] is the radius of the circular path.

At the top of the circle, the tension in the string is zero, so the only force acting on the object is gravity. The gravitational force ([tex]\sf F_g[/tex]) is given by:

[tex]\sf F_g = m \cdot g [/tex]

where:

[tex]\sf g[/tex] is the acceleration due to gravity.

The centripetal force is provided by gravity at the top of the circle, so [tex]\sf F_c = F_g[/tex]:

[tex]\sf \dfrac{m \cdot v^2}{r} = m \cdot g [/tex]

The mass ([tex]\sf m[/tex]) cancels out:

[tex]\sf v^2 = r \cdot g [/tex]

Now, the centripetal acceleration ([tex]\sf a_c[/tex]) is given by:

[tex]\sf a_c = \dfrac{v^2}{r} [/tex]

Substitute the expression for [tex]\sf v^2[/tex]:

[tex]\sf a_c = \dfrac{r \cdot g}{r} [/tex]

[tex]\sf a_c = g [/tex]

Therefore, at the top of the circle, the centripetal acceleration of the weight is equal to [tex]\sf g[/tex].

The correct answer is:

c) Equal to g.